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3 years ago

Major species present when fructose is dissolved in water

Chemistry

2 answers:

Katyanochek1 [597]3 years ago

7 0

The fructose chemical formula is C6H12O6. The answer to the question above regarding the major species present when fructose is dissolved in water (H2O) is "None". No ions are present. It is false that when sugar is dissolved in water there will be strong electrolytes.

BARSIC [14]3 years ago

6 0

Answer:

Alfa-fructofuranose and beta-fructofuranose

Explanation:

The species more stables in water of fructose are their cyclic forms. In aqueous solution, the ketone group of the fructose reacts in a reversible way with the hydroxyl group of the same forming a cyclic hemiketal, in a reaction of intramolecular cycling. These rings formed are more stables when contain 5 or 6 atoms, named furanose and pyranose forms, respectively. In this cycling process the carbonyl carbon (C=O) it is transformed into a chiral center which produces 2 possible anomers, alfa and beta. In the fructose, the hydroxyl group of the formed hemiketal it is produced in the carbon 2 (the anomeric carbon) and can be located above the ring or beneath the ring. In D-sugars (the most common in nature), when the hydroxyl group is located beneath the ring the structure is in the alfa form and when is above the ring is in the beta form.

In the case of fructose dissolved in water it is formed the alfa-fructofuranose an beta-fructofuranose in a majority way, as shown in the picture.

Sugars are capable of suffering of mutarotation when they are dissolved in water, which cause the interconversion of anomers alfa and beta, producing a mix of the alfa and beta forms in the furanose or pyranose rings. And the proportion of each

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Th e enzyme-catalyzed conversion of a substrate at 25°C has a Michaelis constant of 0.045 mol dm−3. Th e rate of the reaction is

Elis [28]

Answer:

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{K_m+[S]}=k_{cat}[E_o]\times \frac{[S]}{K_m+[S]}$

$V_{max}=k_{cat}[E_o]$

v = rate of formation of products =

[S] = Concatenation of substrate

$[K_m]$ = Michaelis constant

$V_{max}$ = Maximum rate achieved

$k_{cat}$ = Catalytic rate of the system

$[E_o]$ = Initial concentration of enzyme

We have :

$K_m=0.045 mol/dm^3$

$v=1.15 mmol/dm^3 s=1.15\times 10^{-3} mol/dm^3 s$

$[S]=0.110 mol/dm^3$

$v=V_{max}\times \frac{[S]}{K_m+[S]}$

$1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}$

$V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s$

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

8 0

3 years ago

The reaction of sodium peroxide and water produces sodium hydroxide and oxygen gas. The following balanced chemical equation rep

a_sh-v [17]

Answer:

3.925 mol.

Explanation:

From the balanced equation:

2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g) ,

It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂ .

Using cross multiplication:

4 moles of NaOH produced with → 1 mole of O₂ .

15.7 moles of NaOH produced with → ??? mole of O₂ .

∴ The no. of moles of O₂ made = (1 mole)(15.7 mole)/(4 mole) = 3.925 mol.

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3 years ago

What happens to temperature and kinetic energy of particles during a change of state

Wittaler [7]

When a substance is changing state, its temperature remains constant. This is because energy is used to increase/decrease kinetic energy of the molecules of the substance, increasing/decreasing the inter-molecular distance and overcoming the energy bonds present between the molecules. Therefore, no energy is used to raise the temperature of the substance and therefore it remains constant

3 0

3 years ago

A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N

vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Initially 0.0733

React 0.0733α α/2 3/2α

Eq 0.0733 - 0.0733α α/2 0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

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3 years ago

What is the new boiling point if 25 g of NaCl is dissolved in 1.0 kg of water

Inessa05 [86]

The correct answer for the question that is being presented above is this one:

Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
delta Tf = Kfm Kf H2O = 1.86 degrees C/m

We need to know the formula for Molality.
molality = mol solute / kg solvent

We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. 

25 g NaCl / 58.5 g/mol = 0.427 mol 

Then, use the formula for molality. 

molality = mol solute / kg solvent 
= 0.427 / 1 
= 0.427 m 

Use now the formula to get the boiling point.

delta Tb = Kbm 
= (0.52)(0.427) 
= 0.22C

8 0

3 years ago